Now, we will turn to the second part of the problem, which is to show that the operator \(\hat w\left( {\bar w} \right)\) is monotone. We will show this in a couple steps. First, note that we can write \({\tilde B_{\bar w}} = B + \left( {N - 1} \right)\bar w\) and for \(e \in E_{\bar w}^{feas}\), \[{C_{\bar w}}\left( e \right) = \mathop {\min }\limits_{w \in W + \left( {1 - 1/N} \right)\bar w} \left\{ {w \cdot \phi \left( e \right)|w \in \partial c\left( e \right)} \right\},\]where \(E_{\bar w}^{feas}\) is an increasing function of \(\bar w\). An increase in \(\bar w\) increases \({\tilde B_{\bar w}}\), and it constrains the cost-minimization problem for implementing \(e\) in two ways: for \(\bar w{ \ge _W}\bar w'\), \(E_{\bar w}^{feas} \subset E_{\bar w'}^{feas}\), and secondly, it impacts the set of feasible contracts for implementing \(e \in E_{\bar w}^{feas}\) at lowest cost and will therefore potentially increase the cost of implementing \(e\). It will be useful to break this analysis up into three steps. First, I will show how an increase in \(\bar w\) impacts the cost-minimization problem. Then, I will argue that any contract in \(\hat w\left( {\bar w} \right)\) is a cost-minimizing contract relative to \(\bar w\). Finally, I will explore how an increase in \(\bar w\) affects \({\tilde B_{\bar w}}\).

When there are only two possible output levels, it is straightforward to characterize the set of cost-minimizing contracts. For all \(e \in {E^{feas}},\) \(w_e^*\left( 0 \right) = 0\), and \(\tilde c{'^ - }\left( e \right) \le w_e^*\left( 1 \right) \le \tilde c{'^ + }\left( e \right)\), where \(\tilde c{'^ - }\left( e \right)\) and \(\tilde c{'^ + }\left( e \right)\) are the left and right derivatives of \(\tilde c\) at \(e\).

We can therefore focus on the much simpler problem of solving for the fixed points of \(\hat e\left( {\bar e} \right).\) Further, \(\hat e\left( {\bar e} \right)\) has some nice properties, which the following proposition shows.

By Topkis's theorem, we therefore have that \(\hat e\left( {\bar e} \right)\) defines a monotone operator on a compact space. The intuition behind this proposition is that, given any cost-minimizing target contract \(w_{\bar e}^*\), each principal either wants to leave \(\left( {1 - 1/N} \right)w_{\bar e}^*\) in place by contributing \(w = 0\), or they want to top up \(\left( {1 - 1/N} \right)w_{\bar e}^*.\) If they choose to top it up, they will top it up to a cost-minimizing contract, which is feasible, because \(w_{\bar e}^*\) is increasing in \(e\). As a result of the monotonicity of \(\hat e\left( {\bar e} \right)\), by an extension of Tarski's fixed-point theorem, there is at least one fixed point, and the set of fixed points forms a complete lattice.

Theorem 2: The set of equilibrium effort levels \({E^*}\) is a complete lattice with least and greatest elements \(e_L^*\) and \(e_H^*\).

Further, the program described by the operator \(\hat e\left( {\bar e} \right)\) can be solved explicitly given any (weakly increasing) cost function \(c\left( e \right)\) and aggregate benefits \(B\). We can therefore ask when it is the case that \(e_L^* < e_H^* < {e^{SB}}\) so that the worst equilibrium effort level is strictly worse than the second-best equilibrium effort level. In particular, given \(B\), I think we will be able to say that for any \(c\) such that \(\tilde c\) has points \(e < e_H^*\) at which it is non-differentiable, for sufficiently large \(N\), then the smallest such point is an equilibrium effort level. This is satisfied, for instance, by \(c\) such that \(c\left( 0 \right) = 0\) and \(c\left( e \right) = F + \frac{c}{2}{e^2},\) in which case \(\tilde c\left( e \right)\) is not differentiable at \(0\). If this conjecture is correct, it is actually nondifferentiability of \(\tilde c\) , rather than nonconvexities in \(c\), that leads to coordination failures. In some cases, such as the example with fixed costs, nonconvexities in \(c\) lead to nondifferentiabilities in \(\tilde c\), but they are neither necessary nor sufficient. They are not necessary, because our initial example with \(E = \left\{ {0,1} \right\}\), \(\tilde c\) is linear (and therefore convex), but it is not differentiable at \(0.\)